Polynomial from Transformed Zeros

Video Explanation

Question

If \( \alpha \) and \( \beta \) are the zeros of the quadratic polynomial

\[ f(x) = x^2 – 2x + 3, \]

find a polynomial whose roots are

\[ \frac{\alpha – 1}{\alpha + 1} \quad \text{and} \quad \frac{\beta – 1}{\beta + 1}. \]

Solution

Step 1: Find \( \alpha + \beta \) and \( \alpha\beta \)

Comparing \(x^2 – 2x + 3\) with \(ax^2 + bx + c\),

\[ a = 1,\quad b = -2,\quad c = 3 \]

\[ \alpha + \beta = -\frac{b}{a} = 2 \]

\[ \alpha\beta = \frac{c}{a} = 3 \]

Step 2: Find the Sum of the New Roots

\[ \frac{\alpha – 1}{\alpha + 1} + \frac{\beta – 1}{\beta + 1} \]

\[ = \frac{(\alpha – 1)(\beta + 1) + (\beta – 1)(\alpha + 1)} {(\alpha + 1)(\beta + 1)} \]

Numerator:

\[ (\alpha\beta + \alpha – \beta – 1) + (\alpha\beta + \beta – \alpha – 1) = 2\alpha\beta – 2 \]

Denominator:

\[ (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 \]

Substitute values:

\[ \text{Sum} = \frac{2(3) – 2}{3 + 2 + 1} = \frac{4}{6} = \frac{2}{3} \]

Step 3: Find the Product of the New Roots

\[ \left(\frac{\alpha – 1}{\alpha + 1}\right) \left(\frac{\beta – 1}{\beta + 1}\right) = \frac{(\alpha – 1)(\beta – 1)} {(\alpha + 1)(\beta + 1)} \]

\[ = \frac{\alpha\beta – (\alpha + \beta) + 1} {\alpha\beta + (\alpha + \beta) + 1} \]

Substitute values:

\[ \text{Product} = \frac{3 – 2 + 1}{3 + 2 + 1} = \frac{2}{6} = \frac{1}{3} \]

Step 4: Form the Required Polynomial

A quadratic polynomial whose zeros have sum \(S\) and product \(P\) is:

\[ x^2 – Sx + P \]

\[ = x^2 – \frac{2}{3}x + \frac{1}{3} \]

Multiplying throughout by 3:

\[ 3x^2 – 2x + 1 \]

Conclusion

The required polynomial is:

\[ \boxed{3x^2 – 2x + 1} \]

\[ \therefore \quad 3x^2 – 2x + 1 \text{ is the required polynomial.} \]